LeetCode – Median of Two Sorted Arrays (Java)

There are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

Java Solution 1

This problem can be converted to the problem of finding kth element, k is (A's length + B' Length)/2.

If any of the two arrays is empty, then the kth element is the non-empty array's kth element. If k == 0, the kth element is the first element of A or B.

For normal cases(all other cases), we need to move the pointer at the pace of half of the array size to get log(n) time.

median-of-two-sorted-array

public double findMedianSortedArrays(int[] nums1, int[] nums2) {
    int total = nums1.length+nums2.length;
    if(total%2==0){
        return (findKth(total/2+1, nums1, nums2, 0, 0)+findKth(total/2, nums1, nums2, 0, 0))/2.0;
    }else{
        return findKth(total/2+1, nums1, nums2, 0, 0);
    }
}
 
public int findKth(int k, int[] nums1, int[] nums2, int s1, int s2){
    if(s1>=nums1.length)
        return nums2[s2+k-1];
 
    if(s2>=nums2.length)
        return nums1[s1+k-1];
 
    if(k==1)
        return Math.min(nums1[s1], nums2[s2]);
 
    int m1 = s1+k/2-1;
    int m2 = s2+k/2-1;
 
    int mid1 = m1<nums1.length?nums1[m1]:Integer.MAX_VALUE;    
    int mid2 = m2<nums2.length?nums2[m2]:Integer.MAX_VALUE;
 
    if(mid1<mid2){
        return findKth(k-k/2, nums1, nums2, m1+1, s2);
    }else{
        return findKth(k-k/2, nums1, nums2, s1, m2+1);
    }
}

Time is O(log(k)).

Java solution 2

Actually, we can write the binary search solution in multiple different ways. The following is one of them.

public static double findMedianSortedArrays(int A[], int B[]) {
	int m = A.length;
	int n = B.length;
 
	if ((m + n) % 2 != 0) // odd
		return (double) findKth(A, B, (m + n) / 2, 0, m - 1, 0, n - 1);
	else { // even
		return (findKth(A, B, (m + n) / 2, 0, m - 1, 0, n - 1) 
			+ findKth(A, B, (m + n) / 2 - 1, 0, m - 1, 0, n - 1)) * 0.5;
	}
}
 
public static int findKth(int A[], int B[], int k, 
	int aStart, int aEnd, int bStart, int bEnd) {
 
	int aLen = aEnd - aStart + 1;
	int bLen = bEnd - bStart + 1;
 
	// Handle special cases
	if (aLen == 0)
		return B[bStart + k];
	if (bLen == 0)
		return A[aStart + k];
	if (k == 0)
		return A[aStart] < B[bStart] ? A[aStart] : B[bStart];
 
	int aMid = aLen * k / (aLen + bLen); // a's middle count
	int bMid = k - aMid - 1; // b's middle count
 
	// make aMid and bMid to be array index
	aMid = aMid + aStart;
	bMid = bMid + bStart;
 
	if (A[aMid] > B[bMid]) {
		k = k - (bMid - bStart + 1);
		aEnd = aMid;
		bStart = bMid + 1;
	} else {
		k = k - (aMid - aStart + 1);
		bEnd = bMid;
		aStart = aMid + 1;
	}
 
	return findKth(A, B, k, aStart, aEnd, bStart, bEnd);
}

Category >> Algorithms

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Bobby Joe Houston

No extra space allowed…
Vaishali Goyal

Can someone explain solution1..how are we dropping the elements?
Stephen Boesch

Complexity is not correct. If k==1 then this O( max(M,N))
Tarun Gulati

What about using a min and a max heap here ? similar to finding the mean of streaming data ?
Udaydeep Thota

For 1st solution, it prints 7 which is correct !!
yuvi

int[] nums1 = new int[]{1,2, 7, 8};
int[] nums2 = new int[]{3, 4, 5, 6, 9, 10, 11, 12, 13};

System.out.println(medianTwoSortedArrays.findMedianSortedArrays(nums1, nums2));
Prints 9.

How is any one of the solutions correct. This makes no sense.
Pushbuffer

Why do you initiate aMid as aMid = aLen * k / (aLen + bLen) ?? What’s the point??
>> Answer is this post:
http://articles.leetcode.com/find-k-th-smallest-element-in-union-of/
Subodh Karwa

If the two arrays are of length ‘n’, the complexity would be O(n), in that case.
The above solution has a better complexity
Yejin Kim

(1) aMid = aLen / 2, (2) k = (aLen + bLen) / 2
2k = (aLen + bLen) which means (3) 2 = (aLen + bLen) / k
then, just substitute (3) to (1)
SM

I used using an array and storing till (n+m)/2. Is this fine.

public class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
int le=nums1.length+nums2.length;
int l =(le)/2;

int[] arr =new int[l+1];
int j=0;
int k=0;
for(int i=0;i=nums1.length || k>=nums2.length){

if(j>=nums1.length){
arr[i]=nums2[k];
k++;
}else{
arr[i]=nums1[j];
j++;
}
}else{
if(nums1[j]nums2[k]){
arr[i]=nums2[k];
k++;
}else{
arr[i]=nums2[k];
if(i+1<=l){
i++;
arr[i]=nums1[j];
j++;
}

k++;
}
}
}
if(le%2==0){
return (arr[l]+arr[l-1])/2.0;
}else
return arr[l];
}
}
Darewreck

I’m not seeing the connection between the two formulas…and how those two is used to derive the other.

aMid = aLen * k/(aLen + bLen)

from
aMid = aLen / 2 and k = （aLen + bLen)/2

at best you get:

aMid = aLen / 2 = (2k-bLen)/2 which is not aLen * k/(aLen + bLen)
Rush

Can you expound on this?
Dean

Thank you. This is the best solution among all online!
toinfinity

Step 5 is correct if and only if length of both array is the same. Only in this occasion, median of two subarrays equal to the median of two original arrays.
Andrew

solution 2 when N>M
public class Solution { public double findMedianSortedArrays(int[] nums1, int[] nums2) { if(nums1.length>nums2.length){ int[] tem= nums1; nums1=nums2; nums2=tem; } int s1 = nums1.length; int s2 = nums2.length; if(s1==0&&s2==0) return 0; int m1=(s1-1)/2; int m2=(s2-1)/2; if(s1==0){ double median= (s2%2==0)? (nums2[m2]+nums2[m2+1])/2.0: nums2[m2]; return median; } double mv1 =0; double mv2=0; mv1= (s1%2==0)? (nums1[m1]+nums1[m1+1])/2.0: nums1[m1]; mv2= (s2%2==0)? (nums2[m2]+nums2[m2+1])/2.0: nums2[m2]; if (mv1==mv2) return mv1; if(s1==1) { if(s2%2==0){ if(nums1[0]nums2[m2+1]) return nums2[m2+1]; else return nums1[0]; } }else{ if (s2==1) return (nums1[0]+nums2[0])/2.0; if(nums1[0]<nums2[m2]){ if (nums1[0]nums2[m2+1]) return (nums2[m2+1]+nums2[m2])/2.0; else return (nums1[0]+nums2[m2])/2.0; } } } int s=s1+s2; if (s % 2 == 0) { if(s1==s2) return findMedianSortedArrays(nums1,0,s1-1,nums2,0,s1-1,false); else{ if(nums1[0]>=nums2[(s2+s1)/2]) return (nums2[(s2+s1)/2-1]+nums2[(s2+s1)/2])/2.0; if(nums1[s1-1]nums1[s1-1])return nums2[(s2-s1)/2]; return findMedianSortedArrays(nums1,0,s1-1,nums2,(s2-s1)/2+1,(s2+s1)/2,true); } } public double findMedianSortedArrays(int[] n1, int nb1,int ne1,int[] n2,int nb2, int ne2, boolean odd) { int s = ne1 - nb1 + 1 ; if(s==2){ if (odd) return Integer.min(Integer.max(n1[nb1],n2[nb2]),Integer.min(n1[ne1],n2[ne2])); return (Integer.max(n1[nb1],n2[nb2])+Integer.min(n1[ne1],n2[ne2]))/2.0; } int mid = (s-1)/2; if (n1[nb1+mid]==n2[nb2+mid]) return n1[nb1+mid]; if (n1[nb1+mid]>n2[nb2+mid]){ if(s%2==0) ne1=nb1+mid+1; else ne1=nb1+mid; return findMedianSortedArrays(n1,nb1,ne1,n2,nb2+ mid,ne2,odd); } if(s%2==0) ne2=nb2+mid+1; else ne2=nb2+mid; return findMedianSortedArrays(n1,nb1+mid,ne1,n2,nb2,ne2,odd); } }
Andrew

when N>M ,tail N to M. then solution 2 O(log m)
Andrew

solution 2 when nM
public static double findMedianSortedArrays(int[] nums1, int[] nums2) {

int s1 = nums1.length;

int s2 = nums2.length;

if(s1==0&&s2==0) return 0;

if(nums1.length>nums2.length){

int[] tem= nums1;

nums1=nums2;

nums2=tem;

}

int m1=(s1-1)/2;

int m2=(s2-1)/2;

if(s1==0)

return (s2%2==0)? (nums2[m2]+nums2[m2+1])/2.0: nums2[m2];

double mv1 =0;

double mv2=0;

mv1= (s1%2==0)? (nums1[m1]+nums1[m1+1])/2.0: nums1[m1];

mv2= (s2%2==0)? (nums2[m2]+nums2[m2+1])/2.0: nums2[m2];

if (mv1==mv2) return mv1;

if (s2==1&&s1==1) return (nums1[0]+nums2[0])/2.0;

if(s1==1) {

if(s2%2==0){

if(nums1[0]nums2[m2+1]) return nums2[m2+1];

else return nums1[0];

}

}else{

if(nums1[0]<nums2[m2]){

if (nums1[0]nums2[m2+1]) return (nums2[m2+1]+nums2[m2])/2.0;

else return (nums1[0]+nums2[m2])/2.0;

}

}

}

int s=s1+s2;

if (s % 2 == 0) {

if(s1==s2)

return findMedianSortedArrays(nums1,0,s1-1,nums2,0,s1-1,false);

else{

if(nums1[0]>=nums2[(s2+s1)/2]) return (nums2[(s2+s1)/2-1]+nums2[(s2+s1)/2])/2.0;

if(nums1[s1-1]nums1[s1-1])return nums2[(s2-s1)/2];

return findMedianSortedArrays(nums1,0,s1-1,nums2,(s2-s1)/2+1,(s2+s1)/2,true);

}

}

public static double findMedianSortedArrays(int[] n1, int nb1,int ne1,int[] n2,int nb2, int ne2, boolean odd) {

int s = ne1 – nb1 + 1 ;

if(s==2){

if (odd) return Integer.min(Integer.max(n1[nb1],n2[nb2]),Integer.min(n1[ne1],n2[ne2]));

return (Integer.max(n1[nb1],n2[nb2])+Integer.min(n1[ne1],n2[ne2]))/2.0;

}

int mid = (s-1)/2;

if (n1[nb1+mid]==n2[nb2+mid]) return n1[nb1+mid];

if (n1[nb1+mid]>n2[nb2+mid]){

if(s%2==0) ne1=nb1+mid+1;

else ne1=nb1+mid;

return findMedianSortedArrays(n1,nb1,ne1,n2,nb2+mid,ne2,odd);

}

if(s%2==0) ne2=nb2+mid+1;

else ne2=nb2+mid;

return findMedianSortedArrays(n1,nb1+mid,ne1,n2,nb2,ne2,odd);

}
Riaz Ud Din

Thanks very much for the explanation. You are a lifesaver. Can you please also explain the logic for int bMid = k – aMid – 1; // b’s middle count
lanoche

in this case a={2,3,4} b={1},i think the answer can not work.
Tajinder Singh

No. It won’t work if the length of the 2 arrays is different.
implementation at:
“http://www.geeksforgeeks.org/median-of-two-sorted-arrays/”
ryanlr

In Java, 1/2*3=0 and 1*3/2=1. Reason is that in the first expression, 1/2 is evaluated as 0.
Yiru

I am not sure why it says index out of bound when I put (aLen/(aLen+bLen)*k) instead of (aLen*k/(aLen+bLen)). Anyone can explain?
Holden

Thank you for the great code. Especially for sweet variables’ names 🙂
(However, explanation of algorithm by Gunner86 is not related to code!)
Holden

Could you please explain those initiating aMid and bMid? I cannot get it.
Holden

Is that wrong to put:
aMid = aLen / 2 and bMid = bLen / 2

Do we get wrong output?
卢刘杰

good idea ! but there is a problem. if we find the kth-1 value, we still can’t decide the kth value in which array.
卢刘杰

because A[aMid] still have possibility to be the kth number. so should include it.
卢刘杰

well done ! but there is a little problem with u method. if aLen=7, bLen=16, k=7, u can get aMid=0. i think you’d better cast the aLen to type double first. this is to say, you’d better write like this
aMid = (int) ( (double)aLen / (aLen + bLen) * k)
卢刘杰

Because in the original post, function findKth is a general function to find kth number in two sorted arrays, not only median of two sorted arrays.
卢刘杰

I suppose your calculation is not right. the result of this algorithm is 6.
卢刘杰

if k==m+n, the program can recursively hit the condition k==0
卢刘杰

scwinji is right. you can read his answer.
卢刘杰

your analysis is more general and awesome and i think you are right. but there is one little problem in your answer. I suppose “If A[aMid] is less than B[bMid], ” shoule be “If A[aMid] is greater than B[bMid], “. anyway , thank you very much.
Jk

I love this site’s solutions, they are so clean and easy to understand! 🙂
bitorange

you can do like this: aMid=(int)(1.0*aLen/(aLen+bLen)*k);
Frank

int aMid = aLen * k / (aLen + bLen);
Frank

why this magic formula? I tried others, but only this one works, whY?
gunners86

when you arrange the two array it would be 25, 28 ,30 ,31. So the median would be average of 28 and 30. 🙂
scwinji

Right, to use the code’s terms, aMid + bMid + 1 = k must be satisfied to be able to make the conclusions it does when A[aMid] > B[bMid] (expanded on later). They also must all be integers >= 0. That line is just how X Wang decided to determine these values, and it has certain runtime properties associated with it. You *could* replace the lines with randomly generating these numbers within bounds:

int aMid, bMid;
Random r = new Random();
if (aLen > bLen) {
bMid = (int)(Math.min(bLen,k) * r.nextDouble());
aMid = k – bMid – 1;
} else {
aMid = (int)(Math.min(aLen,k) * r.nextDouble());
bMid = k – aMid – 1;
}

… which would still get you the right answer, but have pretty different runtime properties.

As for why aMid + bMid + 1 = k is significant: If A[aMid] is less than B[bMid], you know that any elements in after A[aMid] in A can’t be the kth element since there are too many elements in B lower than it (and would exceed k elements). You also know that B[bMid] and any element before B[bMid] in B can’t be the kth element since there are too few elements in A lower than it (there wouldn’t be enough elements before B[bMid] to be the kth element).
scwinji

It’s really odd how this was picked out as the pseudocode to X Wang’s solution, but it doesn’t say anything about the kth element. This algorithm doesn’t have a solution for when an array is of size 1, and also produces an incorrect solution for inputs ar1=[1,2,3] and ar2=[100,100]. On step 4, you get to a point where ar1=[2,3] and ar2=[100,100], and then step 6 calculates the median as (max(2,100) + min(3,100))/2 = 51.5, when it should be actually 3. Sorry, but this is not the same algorithm as the original post.
JY

Which means that you can just do aMid = aLen / 2
Zaga

Also, why int aMid = aLen * k / (aLen + bLen); // a’s middle count

int bMid = k – aMid – 1; // b’s middle count to calculate mid element. Why not aMid = aLen / 2 and bMid = bLen / 2
Zaga

I think when k == m + n is also a special case that should be handled similar to k == 0
Ryan

considering [1,2,3] and [4,5,6,7,8,9,10,11]
the results of this algorithm is 3.5 where the real median is 6. Am i missing something here or is this solution only applys when the two list size are the same?
nisha

What if the two subarrays in the end are {25, 30} and {28, 31}? Here the median should have been average of 25 and 28. This is not attainable going by the logic mentioned by you. Can you please put some light on it?
Shibai Li

Why is this inclusive? Could any one elaborate on this?

aEnd = aMid; // not aEnd = aMid – 1?
Shibai Li

same question here, can anyone give a better answer?
quicker

in the even case you have to run the algo twice. If your findkth func just returns the index and the value, then you can just find the kth -1 value to then find the average without searching for the next value
Vishal

If you merge them and find the median element then time complexity would jump to O(n+m).
Changfeng Chi

大牛
Cheng

Hmm, not really. It should be okay if the length for both arrays or both sub-arrays are different. Open to discuss:)
codingpan

aMid = aLen / 2 and k = （aLen + bLen)/2, so aMid = aLen * k/(aLen + bLen)
Kirk Zhang

I don’t think it matters as long as the invariant i + j = k – 1 is satisfied.
The reason is it might be able to guess the kth element quicker if you weight the arrays.
Kirk Zhang

You have to do it in O(log(length of A + length of B))
javanoob

I might be missing something but since A and B are two sorted arrays, why not just merge them and return the middle element?
mengchao

This is the best solution I have ever seen, simple, efficient and elegant. But I spend a lot of time understanding initiating aMid and bMid. Thank you for sharing this!
Dave

Can you guys tell me if this one is OK ?

public static void main(String[] args) {
int[] arr1 = { 1, 3, 4, 7, 8, 11, 44, 55, 62 };
int[] arr2 = { 2, 4, 5, 7, 33, 56, 77 };
double median = getMedian(arr1, arr2);
System.out.println(“calculated Median : ” + median);
}

private static double getMedian(int[] arr1, int[] arr2) {
int[] medianIndices = getMedianIndices(arr1.length, arr2.length);
double median = 0;
int currIndx1 = 0, currIndx2 = 0, currArr = 0;
if (medianIndices.length == 2) {
for (int i = 0; i <= medianIndices[1]; i++) {
if (arr1[currIndx1] < arr2[currIndx2])
currArr = 1;
else
currArr = 2;

if (i == medianIndices[0] || i == medianIndices[1]) {
if (currArr == 1)
median += arr1[currIndx1];
else
median += arr2[currIndx2];
}

if (currArr == 1)
currIndx1++;
else
currIndx2++;
}
median = median / 2;
} else {
for (int i = 0; i <= medianIndices[0]; i++) {
if (arr1[currIndx1] < arr2[currIndx2])
currArr = 1;
else
currArr = 2;

if (i == medianIndices[0]) {
if (currArr == 1)
median += arr1[currIndx1];
else
median += arr2[currIndx2];
}

if (currArr == 1)
currIndx1++;
else
currIndx2++;
}
}
return median;
}

private static int[] getMedianIndices(int l1, int l2) {
int[] medianIndices;
if ((l1 + l2) % 2 == 0) {
medianIndices = new int[2];
medianIndices[0] = (l1 + l2) / 2 – 1;
medianIndices[1] = (l1 + l2) / 2;
} else {
medianIndices = new int[1];
medianIndices[0] = (l1 + l2) / 2;
}
return medianIndices;
}
n00b23

can anyone answer it?
Jerry Wu

To use this algorithm, Array A and B need to be the same length
gunners86

Perhaps the logic will help to understand it better..

Algorithm:

1) Calculate the medians m1 and m2 of the input arrays ar1[]
and ar2[] respectively.
2) If m1 and m2 both are equal then we are done.
return m1 (or m2)
3) If m1 is greater than m2, then median is present in one
of the below two subarrays.
a) From first element of ar1 to m1 (ar1[0…|_n/2_|])
b) From m2 to last element of ar2 (ar2[|_n/2_|…n-1])
4) If m2 is greater than m1, then median is present in one
of the below two subarrays.
a) From m1 to last element of ar1 (ar1[|_n/2_|…n-1])
b) From first element of ar2 to m2 (ar2[0…|_n/2_|])
5) Repeat the above process until size of both the subarrays
becomes 2.
6) If size of the two arrays is 2 then use below formula to get
the median.
Median = (max(ar1[0], ar2[0]) + min(ar1[1], ar2[1]))/2

Thanks
codergirl227

Kartrace
points out
do the ratio first.
aMid = aLen / (aLen + bLen) * k

NO!!! do the * first. or you will fail. e.g {2, 3, 4} and {1}.

3/4*2=0 3*2/4=1.
Kartrace

do the ratio first.
aMid = aLen / (aLen + bLen) * k
Kun Han

Why do you initiate aMid as aMid = aLen * k / (aLen + bLen) ?? What’s the point??
ryanlr

过奖了，加油。
木木彬

你的算法太厉害了~正在刷leetcode
online chesstle

the edge conditions are incorrect start + k is not the kth element. For Example when start==0 and we are looking to get 8th element, we would end up with array[8] whereas we need array[7] (th: // Handle special cases
if (aLen == 0)
return B[bStart + k];
if (bLen == 0)
return A[aStart + k];
ecc

I have never used java until today but when I tried this code, I noticed that this line:
int aMid = aLen * k / (aLen + bLen);
overflows very easily when aLen, k > 65536
I used a very ugly hack:
int aMid = k * 4 / (aLen + bLen) * aLen / 4
But what is the best practice?

LeetCode – Median of Two Sorted Arrays (Java)

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